LADR Exercises 1A

Problem 1

Show that $\alpha +\beta =\beta +\alpha$ for all $\alpha ,\beta \in \mathbb{C}$.

Proof. Let $\alpha =a+bi$, $\beta =c+di$ for $a,b,c,d\in \mathbb{R}$. Then, we have:

$$\begin{array}{rl}\alpha +\beta & =(a+bi)+(c+di)\\ & =(a+c)+(b+d)i\end{array}$$

And also:

$$\begin{array}{rl}\beta +\alpha & =(c+di)+(a+bi)\\ & =(a+c)+(d+b)i\end{array}$$

Since addition of real numbers is commutative, we have $a+c=c+a$ and $b+d=d+b$, so

$$\begin{array}{rl}(a+c)+(b+d)i& =(c+a)+(d+b)i\\ \alpha +\beta & =\beta +\alpha \end{array}$$

as desired.

Problem 2

Show that $(\alpha +\beta )+\lambda =\alpha +(\beta +\lambda )$ for all $\alpha ,\beta ,\lambda \in \mathbb{C}$.

Proof. Let $\alpha =a_1+a_{2}i$, $b=b_1+b_{2}i$, and $\lambda =c_1+c_{2}i$, for $a_k,b_k,c_k\in \mathbb{R}$. Then:

$$\begin{array}{rl}(\alpha +\beta )+\lambda & =[(a_1+a_{2}i)+(b_1+b_{2}i)]+(c_1+c_{2}i)\\ & =[(a_1+b_1)+(a_2+b_2)i]+(c_1+c_{2}i)\\ & =[(a_1+b_1)+c_1]+[(a_2+b_2)+c_2]i\\ & =[a_1+(b_1+c_1)]+[a_2+(b_2+c_2)]i\\ & =(a_1+a_{2}i)+[(b_1+c_1)+(b_2+c_2)i]\\ & =(a_1+a_{2}i)+[(b_1+b_{2}i)+(c_1+c_2)i]\\ & =\alpha +(\beta +\lambda )\end{array}$$

as desired.

Problem 3

Show that $(\alpha \beta )\lambda =\alpha (\beta \lambda )$ for all $\alpha ,\beta ,\lambda \in \mathbb{C}$.

This proof feels clumsy. Maybe there’s a better way to do this?

Proof. Let $\alpha =a_1+a_{2}i$, $b=b_1+b_{2}i$, and $\lambda =c_1+c_{2}i$, for $a_k,b_k,c_k\in \mathbb{R}$. Then:

$$\begin{array}{rl}\alpha \beta & =(a_1+a_{2}i)(b_1+b_{2}i)\\ & =a_1{b}_1+a_1{b}_{2}i+a_2{b}_{1}i+a_2{b}_2{i}^2\\ & =(a_1{b}_1-a_2{b}_2)+(a_1{b}_2+a_2{b}_1)i\\ & =x+yi\end{array}$$

where $x=a_1{b}_1-a_2{b}_2$, $y=a_1{b}_2+a_2{b}_1$. Then, $(\alpha \beta )\lambda$ is:

$$\begin{array}{rl}\alpha \beta (\lambda )& =(x+yi)(c_1+c_{2}i)\\ & =x{c}_1+x{c}_{2}i+y{c}_{1}i+y{c}_2{i}^2\\ & =(x{c}_1-y{c}_2)+(x{c}_2+y{c}_1)i\end{array}$$

For $\beta \lambda$, we have:

$$\begin{array}{rl}\beta \lambda & =(b_1+b_{2}i)(c_1+c_{2}i)\\ & =b_1{c}_1+b_1{c}_{2}i+b_2{c}_{1}i+b_2{c}_2{i}^2\\ & =(b_1{c}_1-b_2{c}_2)+(b_1{c}_2+b_2{c}_1)i\\ & =p+qi\end{array}$$

where $p=b_1{c}_1-b_2{c}_2$, $q=b_1{c}_2+b_2{c}_1$. Then, $\alpha (\beta \lambda )$ is:

$$\begin{array}{rl}\alpha (\beta \lambda )& =(a_1+a_{2}i)(p+qi)\\ & =a_{1}p+a_{1}qi+a_{2}pi+a_{2}q{i}^2\\ & =(a_{1}p-a_{2}q)+(a_{1}q+a_{2}p)i\end{array}$$

Expanding our simplifications:

$$\begin{array}{rl}\alpha \beta (\lambda )& =(x{c}_1-y{c}_2)+(x{c}_2+y{c}_1)i\\ & =(a_1{b}_1{c}_1-a_2{b}_2{c}_1-a_1{b}_2{c}_2-a_2{b}_1{c}_2)+(a_1{b}_1{c}_2-a_2{b}_2{c}_2+a_1{b}_2{c}_1+a_2{b}_1{c}_1)i\end{array}$$

and

$$\begin{array}{rl}\alpha (\beta \lambda )& =(a_{1}p-a_{2}q)+(a_{1}q+a_{2}p)i\\ & =(a_1{b}_1{c}_1-a_1{b}_2{c}_2-a_2{b}_1{c}_2-a_2{b}_2{c}_1)+(a_1{b}_1{c}_2+a_1{b}_2{c}_1+a_2{b}_1{c}_1-a_2{b}_2{c}_2)i\\ & =(a_1{b}_1{c}_1-a_2{b}_2{c}_1-a_1{b}_2{c}_2-a_2{b}_1{c}_2)+(a_1{b}_1{c}_2-a_2{b}_2{c}_2+a_1{b}_2{c}_1+a_2{b}_1{c}_1)i\\ & =a(\beta \lambda )\end{array}$$

as desired.

Problem 4

Show that $\lambda (\alpha +\beta )=\lambda \alpha +\lambda \beta$ for all $\alpha ,\beta ,\lambda \in \mathbb{C}$.

Proof. Let $\alpha =a+bi,\beta =c+di$ and $\lambda =e+fi$, where $a,b,c,d,e,f\in \mathbb{R}$. Then:

$$\begin{array}{rl}\lambda (\alpha +\beta )& =(e+fi)[(a+bi)+(c+di)]\\ & =(e+fi)[(a+c)+(b+d)i]\\ & =e(a+c)+e(b+d)i+fi(a+c)+fi(b+d)i\\ & =ea+ec+ebi+edi+fai+fci-fb-fd\\ & =(ea+ec-fd-fb)+(af+eb+ed+cf)i\end{array}$$

and

$$\begin{array}{rl}\lambda (\alpha )+\lambda \beta & =(e+fi)(a+bi)+(e+fi)(c+di)\\ & =(ea+afi+ebi-fb)+(ec+cfi+edi-fd)\\ & =(ea+ec-fd-fb)+(af+eb+cf+ed)i\end{array}$$

Thus, we have $\lambda (\alpha +\beta )=\lambda (\alpha )+\lambda (\beta )$ as desired.

Problem 5

Show that for every $\alpha \in \mathbb{C}$, there exists a unique $\beta \in \mathbb{C}$ such that $\alpha +\beta =0$.

Proof. Let $\alpha =a_1+a_{2}i$, where $a_k\in \mathbb{R}$, and $\beta =-a_1-a_{2}i$. Then, we have:

$$\begin{array}{rl}\alpha +\beta & =(a_1+a_{2}i)+(-a_1-a_{2}i)\\ & =(a_1-a_1)+(a_2-a_2)i\\ & =0+0i\\ & =0\end{array}$$

which proves existence. To show that $\beta$ is unique, suppose $\lambda \in \mathbb{C}$ such that $\alpha +\lambda =0$. Then:

$$\begin{array}{rl}\lambda & =\lambda +(\alpha +\beta )\\ & =(\lambda +\alpha )+\beta \\ & =0+\beta \\ & =\beta \end{array}$$

and thus $\beta$ is unique.

Problem 6

Show that for every $\alpha \in \mathbb{C}$ with $\alpha \ne 0$, there exists a unique $\beta \in \mathbb{C}$ such that $\alpha \beta =1$.

Proof. Let $\alpha =a+bi$ and $\beta =\frac{1}{a+bi}$. Then, we have:

$$\begin{array}{rl}\alpha \beta & =(a+bi)\left(\frac{1}{a+bi}\right)\\ & =1\left(\frac{a+bi}{a+bi}\right)\\ & =1(1)\\ & =1\end{array}$$

which proves existence. To show $\beta$ is unique, suppose $\lambda \in \mathbb{C}$ such that $\alpha \lambda =1$. Then:

$$\begin{array}{rl}\lambda & =\lambda (\alpha \beta )\\ & =(\lambda \alpha )(\beta )\\ & =(1)(\beta )\\ & =\beta \end{array}$$

and thus $\beta$ is unique.

Problem 7

Show that

$$\frac{-1+\sqrt{3}i}{2}$$

is a cube root of $1$ (meaning its cube equals $1$).

Proof. We have

$${\left(\frac{-1+\sqrt{3}i}{2}\right)}^2=\frac{1-\sqrt{3}i-\sqrt{3}i-3}{4}=\frac{-2-2\sqrt{3}i}{4}=\frac{-1-\sqrt{3}i}{2}$$

Then:

$$\begin{array}{rl}{\left(\frac{-1+\sqrt{3}i}{2}\right)}^3& ={\left(\frac{-1+\sqrt{3}i}{2}\right)}^2\left(\frac{-1+\sqrt{3}i}{2}\right)\\ & =\left(\frac{-1-\sqrt{3}i}{2}\right)\left(\frac{-1+\sqrt{3}i}{2}\right)\\ & =\frac{1-\sqrt{3}i+\sqrt{3}i-\sqrt{3}\sqrt{3}{i}^2}{4}\\ & =\frac{1-(3)(-1)}{4}\\ & =\frac{4}{4}=1\end{array}$$

as desired.

Problem 8

Find two distinct square roots of $i$.

Suppose $a,b\in \mathbb{R}$ for some $a+bi$ such that $(a+bi{)}^2=i$. Then, we have:

$$(a+bi{)}^2=(a+bi)(a+bi)=a^2+2abi-b^2=i$$

Grouping real and imaginary terms together, we have

$$(a^2-b^2)+(2ab)i=i$$

Since the real and imaginary parts on each side must be equal, we have

$$\begin{array}{rl}{a}^2+b^2& =0\\ ab& =\frac{1}{2}\end{array}$$

The first equation tells us that $b=\pm a$. If we have $b=-a$, we would have

$$a(-a)=-a^2=\frac{1}{2}$$

which is impossible since $a,b\in \mathbb{R}$. Thus, we must have $b=a$, which gives:

$$\begin{array}{rl}{a}^2& =\frac{1}{2}\\ a& =\pm \frac{1}{\sqrt{2}}\end{array}$$

Hence, our two roots are:

$$a+bi=\pm \left(\frac{1}{\sqrt{2}}\right)(1+i)$$

as desired.

Problem 9

Find $x\in {\mathbb{R}}^4$ such that

$$(4,-3,1,7)+2x=(5,9,-6,8)$$

We have:

$$\begin{array}{rl}2x& =(5,9,-6,8)-(4,-3,1,7)\\ & =(1,12,-7,1)\end{array}$$

which then gives:

$$x=\frac{1}{2}(1,12,-7,1)=\left(\frac{1}{2},6,-\frac{7}{2},\frac{1}{2}\right)$$

Problem 10

Explain why there does not exist $\lambda \in \mathbb{C}$ such that

$$\lambda (2-3i,5+4i,-6+7i)=(12-5i,7+22i,-32-9i)$$

Proof. If such $\lambda \in \mathbb{C}$ exists, then we have

$$\begin{array}{rl}\lambda (2-3i)& =12-5i\\ \lambda (-6+7i)& =-32-9i\end{array}$$

Then, we must have:

$$\begin{array}{rl}\lambda (2-3i)(-32-9i)& =\lambda (-6+7i)(12-5i)\\ -91+78i& =-37+114i\end{array}$$

which is impossible. Hence, such $\lambda \in \mathbb{C}$ does not exist.

Problem 11

Show that for every $(x+y)+z=x+(y+z)$ for all $x,y,z\in {\mathbb{F}}^n$.

Proof. We have $x=(x_1,\dots ,x_n)$, $y=(y_1,\dots ,y_n)$ and $z=(z_1,\dots ,z_n)$. Then:

$$\begin{array}{rl}(x+y)+z& =((x_1,\dots ,x_n)+(y_1,\dots ,y_n))+(z_1,\dots ,z_n)\\ & =(x_1+y_1,\dots ,y_1+y_n)+(z_1,\dots ,z_n)\\ & =((x_1+y_1)+z_1,\dots ,(x_n+y_n)+z_n)\\ & =(x_1+(y_1+z_1),\dots ,x_n+(y_n+z_n))\\ & =(x_1,\dots ,x_n)+(y_1+z_1,\dots ,y_n+z_n)\\ & =(x_1,\dots ,x_n)+((y_1,\dots ,y_n)+(z_1,\dots ,z_n))\\ & =x+(y+z)\end{array}$$

Problem 12

Show that for every $(ab)x=a(bx)$ for all $x\in {\mathbb{F}}^n$ and all $a,b\in \mathbb{F}$.

Proof. We have $x=(x_1,\dots ,x_n)$ and $y=(y_1,\dots ,y_n)$. Then:

$$\begin{array}{rl}ab(x)& =ab(x_1,\dots ,x_n)\\ & =(ab(x_1),\dots ,(ab)x_n)\\ & =(a(b{x}_1),\dots ,a(b{x}_n))\\ & =a(b{x}_1,\dots ,b{x}_n)\\ & =a(bx)\end{array}$$

Problem 13

Show that $1x=x$ for all $x\in {\mathbb{F}}^n$.

Proof. We have $x=(x_1,\dots ,x_n)$. Then

$$1x=1(x_1,\dots ,x_n)=(1\cdot x_1,\dots ,1\cdot x_n)=(x_1,\dots ,x_n)=x$$

Problem 14

Show that $\lambda (x+y)=\lambda x+\lambda y$ for all $\lambda \in \mathbb{F}$ and all $x,y\in \mathbb{F}$.

Proof. We have $x=(x_1,\dots ,x_n)$ and $y=(y_1,\dots ,y_n)$. It follows that:

$$\begin{array}{rl}\lambda (x+y)& =\lambda ((x_1,\dots ,x_n)+(y_1,\dots ,y_n))\\ & =\lambda ((x_1+y_1)+\cdots +(x_n+y_n))\\ & =(\lambda (x_1+y_1)+\cdots +\lambda (x_n+y_n))\\ & =((\lambda x_1+\cdots +\lambda x_n))+(\lambda y_1+\cdots +\lambda y_n)\\ & =\lambda x+\lambda y\end{array}$$

Problem 15

Show that $(a+b)x=ax+bx$ for all $a,b\in \mathbb{F}$ and all $x\in {\mathbb{F}}^n$.

Proof. Suppose $x=(x_1,\dots ,x_n)$. Then

$$\begin{array}{rl}(a+b)x& =(a+b)(x_1,\dots ,x_n)\\ & =(a{x}_1+b{x}_1,\dots ,a{x}_n+b{x}_n)\\ & =(a{x}_1,\dots a{x}_n)+(b{x}_1,\dots ,b{x}_n)\\ & =a(x_1,\dots ,x_n)+b(x_1,\dots ,x_n)\\ & =ax+bx\end{array}$$